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已知函数F(x)=2AsinxCosx+√3Cos²x%√3sin&#178...

f(x)=cosx(sinxcosπ/3+cosxsinπ/3)-√3cos²x+√3/4 =(1/2)sinxcosx+(√3/2)cos²x-√3cos²x+√3/4 =(1/2)sinxcosx-(√3/2)cos²x+√3/4 =(1/4)sin2x-(√3/4)cos2x =(1/2)sin(2x-π/3) 所以T=π x∈[-π/4,π/4]时, 则2x-π/3∈[-5π/6,π/6] ...

f(x)=sin2x+cos2x+2 =√2sin(2x+π/4)+2 (1) 2kπ-π/2

解: (1) f(x)=2√3sinxcosx+2cos²x-t =√3sin(2x)+1+cos(2x)-t =√3sin(2x)+cos(2x)+1-t =2[(√3/2)sin(2x)+(1/2)cos(2x)]+1-t =2sin(2x+π/6)+1-t f(x)=0 2sin(2x+π/6)+1-t=0 sin(2x+π/6)=(t-1)/2 x∈[0,π/2],则π/6≤2x+π/6≤7π/6 -½≤sin(...

f(x)=2cosx(sinx/2+√3cosx/2)-√3sin²x+sinxcosx =2sinxcosx+√3(cos²x-sin²x) =sin2x+√3cos2x =2(sin2x/2+√3cos2x/2) =2sin(2x+π/3); 则最小正周期 T=2π÷2=π

解: f(x)=sin²x+2sinxcosx+3cos²x =sin²x+cos²x+2cos²x+sin2x =1+2cos²x+sin2x =1+con2x+1+sin2x =2+√2sin(π/4+2x) ∵|sin(π/4+2x)|0 ∵正弦函数的周期是2π ∴sin(π/4+2x)的周期是π 因为f(x)>0 ∴最小正周期是π...

由余弦定理cosx=(a²+c²-b²)/(2ac) =(a²+c²-ac)/(2ac) =(a²+c²)/(2ac)-1/2 ≥2ac/(2ac)-1/2 =1/2 所以0

f(x)=sin²x+2√3sinxcosx-cos²x = 2√3sinxcosx-(cos²x-sin²x) = √3sin2x-cos2x = 2(sin2xcosπ/6-cos2xsinπ/6) = 2sin(2x-π/6) 当2x-π/6=2kπ+π/2时,sin(2x-π/6)有最大值2,此时2x=2kπ+2π/3,x=kπ+π/3,其中k∈Z (2x-π/6)∈(...

1、 向量m·向量n=2cos²x+2√3sinxcosx - y=0 y=2cos²x+2√3sinxcosx =2cos²x - 1 + 2√3sinxcosx + 1 =cos2x + √3sin2x + 1 =2[(1/2)cos2x+(√3/2)sin2x] + 1 =2sin(2x + π/6) + 1 T=2π/2=π 2、 f(A/2)=2sin(A + π/6) + 1=3, sin(A +...

1)∵sinx+cosx=1/5 ∴(sinx+cosx)²=1/25 ∴sinx²+2sinxcosx+cosx²=1/25 ∵sinx²+cosx²=1 ∴2sinxcosx=1/25-1=-24/25 sinxcosx=12/25 ∴sinx²-2sinxcosx+cosx²=1+24/25=49/25 ∴(sinx-cosx)²=49/25 ∴sinx-c...

y = sin²x+2sinxcosx+3cos²x = 2sinxcosx+2cos²x +cos²x+sin²x = sin2x+cos2x+1+1 = √2sin(2x+π/4)+2 值域【2-√2,2+√2】

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